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Question 14 in F-I-S section 1.2 asks:

Let $\mathbf{V}=\{(a_1,a_2,\ldots ,a_n)\colon a_i\in \mathbb{C}$ for $i=1,2,\ldots n\}$; so $\mathbf{V}$ is a vector space over $\mathbb{C}$. Is $\mathbf{V}$ a vector space over the field of real numbers with the operations of coordinatewise addition and multiplication?

My question concerns which field operators are relevant when considering $\mathbf{V}$ over $\mathbb{R}$. I'm confused because if we take $x=(a+bi)$ and $y=(c+di)$ where $b,d\neq 0$ so $x,y\in\mathbf{V}$, then being "over the field of $\mathbb{R}$" would be create a conflict when considering $x+y$ (the reasoning being that complex addition is not defined for the real numbers).

Even with scalar multiplication, if we multiply $x\in\mathbf{V}$ by $a\in \mathbb{R}$, then I'm still using scalar multiplication as defined in the field $\mathbb{C}$.

So why if I'm taking $\mathbf{V}$ over $\mathbb{R}$ are we still using the field operators as defined by $\mathbb{C}$?

EDIT:

Thank you for the responses! But I think I'm still confused by the fact that $\mathbf{V}$ is a set of vectors with complex components but it's being taken over $\mathbb{R}$.

So let's say I'm now checking to see if $\mathbf{V}$ is a vector space over $\mathbb{R}$. If I take this set $\mathbf{V}$ over $\mathbb{R}$, still retaining coordinatewise addition, then when I perform $(a_1,a_2,\ldots ,a_n)+(b_1,b_2,\ldots ,b_n)$ with $a_i,b_i\in\mathbb{C}$, then $a_i+b_i$ (having complex parts) isn't defined with the addition operator for $\mathbb{R}$. So in this case, how do I verify the vector space requirements such as associativity if addition (from $\mathbb{R}$) isn't defined for the vectors in $\mathbf{V}$ (those that have components with complex parts).

Or is it true to say that the addition operator from the field $\mathbb{R}$ is still defined when adding complex numbers?

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  • $\begingroup$ You defined $\;V\;$ as a set of $\;n$- vectors, and thus $\;x,y\notin V\;$, unless $\;n=1\;$ . After this is understood and fixed, then yes: the operation of multiplication by scalar in each coordinate of each element of $\;V\;$ is exactly the same as the multiplication within $\;\Bbb C\;$, but you only use real scalars. Perhaps later you could check that $\;\dim_{\Bbb R}V=2n\;$ $\endgroup$ – Timbuc Nov 21 '14 at 16:58
  • $\begingroup$ Thanks for your response, @Timbuc! I understood your response but I still have some confusion. I edited my post in order to clarify what is causing my confusion. In general, I think I don't understand the semantics/terminology regarding the phrase "... vector space over field ..." in terms of the chosen operators. In this case especially because the vector space of interest is a set of complex $n$-tuples, but then it is taken "over" the reals. $\endgroup$ – David Gong Nov 21 '14 at 19:16
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In case $n=1$ the problem is to show that $\mathbb{C}$ is a vector space over $\mathbb{R}$. Check the axioms: write $a+bi$ as $(a,b)$, and then all of the verifications are identical to those for $\mathbb{R}^2$. You see that the complex field operations don't come in to the question at all: rather, some vector which are dependent over $\mathbb{C}$ become independent over $\mathbb{R}$, most obviously $1$ and $i$.

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  • $\begingroup$ Thank you for your response! I edited my post to clarify my confusion about but your answer helps. I just want to confirm what I'm thinking, regarding the textbook question specifically, is it okay to treat $a+bi$ as $(a,b)$ even though $\mathbf{V}$ is already a clearly defined set? In which case I wonder if it is important to note the equivalence of $a+bi$ and $(a,b)$ which I know seems obvious but isn't entirely clear to me. If this is the case then I believe the problem becomes showing that the vector space of $n$ ordered pairs in $\mathbb{R}\times\mathbb{R}$ is still a vector space? $\endgroup$ – David Gong Nov 21 '14 at 19:28
  • $\begingroup$ What I'm doing is essentially assuming that $\mathbb{C}$ and $\mathbb{R}^2$ are isomorphic vector spaces. There's no need to do this: just point out that $\mathbb{C}$ is an abelian group, that for $r,s$ real and $z, w$ complex, $r(sz)=rs(z), r(z+w)=rz+rw,(r+s)z=rz+sz$. In any case, if you do want to use this isomorphism then you're right that you need to justify it, and if you haven't heard the word isomorphism yet then you probably don't have the tools to do so rigorously (if you have heard the isomorphism, then just prove this is one!) But given this, the problem transforms as you say. $\endgroup$ – Kevin Arlin Nov 21 '14 at 20:27
  • $\begingroup$ In response to your edit: read the axioms of a vector space more carefully. The addition on a vector space has nothing, a priori, to do with the addition on its field of scalars-it's just a commutative, associative operation with an identity and inverses! So it's irrelevant that the addition in $\mathbb{C}^n$ is "not defined over $\mathbb{R}$." $\endgroup$ – Kevin Arlin Nov 21 '14 at 20:29
  • $\begingroup$ Your last comment is probably what helped the most. I'm actually just studying for a final so I'm trying to solidify my knowledge of these fundamentals at this point which is why I'm being so nitpicky. One last question: I can't seem to convince myself that addition on a vector space has nothing to do with addition on its field of scalars. At least in this case, the vector space's addition is "coordinatewise". And since the vector coordinates are from $\mathbb{C}$ isn't it actually relevant that the field $\mathbb{R}$? this really is my last question, I swear! $\endgroup$ – David Gong Nov 21 '14 at 21:01

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